Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 918

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\[1)\ y = \sqrt{2 - 3x^{2}}\]

\[Пусть\ u = 2 - 3x^{2};\ y(u) = \sqrt{u}:\]

\[y^{'}(x) = \left( 2 - 3x^{2} \right)^{'} \bullet \left( \sqrt{u} \right)^{'};\]

\[y^{'}(x) = ( - 3 \bullet 2x) \bullet \frac{1}{2\sqrt{u}} =\]

\[= - \frac{3x}{\sqrt{2 - 3x^{2}}}.\]

\[Точки\ экстремума:\]

\[- 3x = 0\ \]

\[x = 0.\]

\[Выражение\ имеет\ смысл\ при:\]

\[2 - 3x^{2} \geq 0\]

\[3x^{2} \leq 2\]

\[x^{2} \leq \frac{2}{3}\]

\[- \frac{\sqrt{6}}{3} < x < \frac{\sqrt{6}}{3}.\]

\[Ответ:\ \ x_{1} = - \frac{\sqrt{6}}{3};\ \ x_{2} = 0;\ \ \]

\[x_{3} = \frac{\sqrt{6}}{3}.\]

\[2)\ y = \sqrt{x^{3} - 3x}\]

\[Пусть\ u = x^{3} - 3x;\ \ y(u) = \sqrt{u}:\]

\[y^{'}(x) = \left( x^{3} - 3x \right)^{'} \bullet \left( \sqrt{u} \right);\]

\[y^{'}(x) = \left( 3x^{2} - 3 \right) \bullet \frac{1}{2\sqrt{u}} =\]

\[= \frac{3}{2} \bullet \frac{x^{2} - 1}{\sqrt{x^{3} - 3x}}.\]

\[Точки\ экстремума:\]

\[x^{2} - 1 = 0\]

\[x^{2} = 1\]

\[x = \pm 1.\]

\[Выражение\ имеет\ смысл\ при:\]

\[x^{3} - 3x \geq 0\]

\[x \bullet \left( x^{2} - 3 \right) \geq 0\]

\[\left( x + \sqrt{3} \right) \bullet x \bullet \left( x - \sqrt{3} \right) \geq 0\]

\[- \sqrt{3} \leq x \leq 0\ или\ x \geq \sqrt{3}.\]

\[Ответ:\ \ x_{1} = - \sqrt{3};\ \ x_{2} = - 1;\ \ \]

\[x_{3} = 0;\ \ x_{4} = \sqrt{3}.\]

\[3)\ y = |x - 1|\]

\[y^{'}(x) = \pm (x - 1)^{'} = \pm (1 - 0) =\]

\[= \pm 1.\]

\[Точка\ излома\ графика:\]

\[x - 1 = 0\ \]

\[x = 1.\]

\[Ответ:\ \ x = 1.\]

\[4)\ y = x^{2} - |x| - 2\]

\[y^{'}(x) = \left( x^{2} \right)^{'} \pm (x)^{'} - (2)^{'} =\]

\[= - 2x \pm 1.\]

\[Точки\ экстремума:\]

\[- 2x \pm 1 = 0\]

\[- 2x = \pm 1\]

\[x = \pm \frac{1}{2}.\]

\[Точка\ излома\ графика:\]

\[x = 0.\]

\[Ответ:\ \ x_{1} = - \frac{1}{2};\ \ x_{2} = 0;\ \ \]

\[x_{3} = \frac{1}{2}.\]